PROFESSOR TODA: And Calc II.
And I will go ahead and solve some problems today out
of chapter 10 as a review.
Meaning what?
Meaning, that you have section 10.1 followed by 10.2
followed by 10.4.
These ones are required sections,
but I'm putting the material all together as a compact set.
So, if we cannot officially cut between, as I told you,
cut between the sections.
One thing that I did not work examples on,
trusting that you'd remember it was integration.
In particular, I didn't cover integration
of vector valued functions and examples that
are very very important.
Now, do you need to learn something special for that?
No.
But just like you cannot learn organic chemistry without
knowing inorganic chemistry, then you could not know how
to integrate a vector value function r prime of d to get r
of d unless you know calculus one and caluculus two, right?
So let's say first a bunch of formulas
that you use going back to last week's knowledge
what have we learned?
We work with regular curves in r3.
And in particular if they are part of R2,
they are plain curves.
I want to encourage you to ask questions
about the example [INAUDIBLE] now.
In the review session we have applications [INAUDIBLE]
from 2 2 3.
What was a regular curve?
Is anybody willing to tell me what a regular curve was?
Was it vector value function?
Do you like big r or little r?
STUDENT: Doesn't matter.
PROFESSOR TODA: Big r of t.
Vector value function.
x of t [INAUDIBLE] You know, I told you that sometimes we
use brackets here.
Sometimes we use round parentheses depending
how you represent a vector in r3 in our book they use brackets,
but in other calculus books, they use round parentheses
around it.
So these are the coordinates of the moving particle in time.
Doesn't have to be a specific object, could be a fly,
could be just a particle, anything
in physical motion between this point a of b equals a and b
of t equals b.
So at time a and time b you are there.
What have we learned?
We've learned that a regular curve means its differentiable
and the derivative is continuous, it's a c1 function.
And what else?
The derivative of the position vector
called velocity never vanishes.
So it's different from 0 for every t in the interval
that you take, like ab.
That's a regular curve.
Regular curve was something we talked about at least 5 times.
The point is how do we see the backwards process?
That means if somebody gives you the velocity of a vector curve,
they ask you for the position vector.
So let's see an example.
Integration example 1 says I gave you
the veclocity vector or a certain law of motion
that I don't know.
I just know the velocity vector is being 1 over 1
plus t squared.
Should I put the brace here?
An angular bracket?
One over one plus t squared.
And I'm gonna put a cosign on 2t, and t squared
plus equal to minus t.
And somebody says, that's all I know for P
on an arbitrary real integral.
And we know via the 0 as being even.
Let's say it's even as 0 0 and that
takes a little bit of thinking.
I don't know.
How about a 1, which would be just k.
Using this velocity vector find me being normal,
which means find the position vector
corresponding to this velocity.
What is this?
It's actually initial value
STUDENT: [INAUDIBLE] 1, 1, and 1?
PROFESSOR TODA: 0, what is it?
When place 0 in?
STUDENT: Yeah.
[INTERPOSING VOICES]
STUDENT: Are these the initial conditions
for the location, or--
PROFESSOR TODA: I'm sorry.
I wrote r the intial condition for the location.
Thank you so much, OK?
I probably would've realized it as soon as possible.
Not the initial velocity I wanted to give you,
but the initial position.
All right, so how do I get to the r of d?
I would say integrate, and when I integrate,
I have to keep in mind that I have to add the constants.
Right?
OK.
So from v, v is our priority.
It follows that r will be-- who tells me?
Do you guys remember the integral of 1 plus t squared?
STUDENT: [INAUDIBLE]
PROFESSOR TODA: So that's the inverse.
Or, I'll write it [? arc tan, ?] and I'm very happy that you
remember that, but there are many students who don't.
If you feel you don't, that means that you have to open
the -- where? -- Between chapters 5 and chapter 7.
You have all these integration chapters--
the main ones over there.
It's a function definted on the whole real interval,
so I don't care to worry about it.
This what we call an IVP, initial value problem.
So what kind of problem is that?
It's a problem like somebody would
give you knowing that f prime of t is the little f,
and knowing that big f of 0 is the initial value
for your function of find f.
So you have actually an initial value problem of the calc
that you've seen in previous class.
arctangent of t plus c1 and then if you miss the c1 in general,
this can mess up the whole thing because-- see, in your case,
you're really lucky.
If you plug in the 0 here, what are you gonna have?
You're gonna have arctangent of 0, and that is 0.
So in that case c1 is just 0.
And [? three ?] [? not ?] and if you forgot it would not be
the end of the world, but if you forgot it in general,
it would be a big problem.
So don't forget about the constant.
When you integrate-- the familiar of antiderivatives
is cosine 2t.
I know you know it.
1/2 sine of t.
Am I done?
No, I should say plus C2.
And finally the familiar of antiderivatives of t
squared plus e to minus t.
STUDENT: 2t minus e to the negative t.
PROFESSOR TODA: No, integral of.
So what's the integral of--
STUDENT: t 2 squared.
PROFESSOR TODA: t cubed over 3-- minus, excellent.
Now, do you want one of you guys almost
kill me during the weekend.
But that's OK.
I mean, this problem had something
to do with integral minus.
He put that integral of e to the minus t was equal to minus t.
So pay attention to the sign.
Remember that integral of e to the at,
the t is to the at over a plus.
Right?
OK, so this is what you have, a minus plus C3.
Pay attention also to the exam.
Because in the exams, when you rush,
you make lots of mistakes like that.
R of 0 is even.
So the initial position is given as C1.
I'm replacing in my formula.
It's going to be C1, C2, and what?
When I replace the 0 here, what am I going to get?
STUDENT: You're going to get negative 1.
PROFESSOR TODA: Minus 1 plus C3.
Note that I fabricated this example, so that C3 is not
going to be 0.
I wanted some customs to be zero and some customs
to not be 0, just for you to realize it's
important to pay attention.
OK, minus 1 plus C3.
And then I have 0, 0, 1 as given as initial position.
So what do you get by solving this linear system that's
very simple?
In general, you can get more complicated stuff.
C1 is 0, C2 is 0, C3 is a--
STUDENT: 2.
PROFESSOR TODA: 2.
And so it was a piece of cake.
What is my formula?
If you leave it like that, generally you're
going to get full credit.
What would you need to do to get full credit?
STUDENT: Rt is equal to R10 plus 1/2 sine of 2t plus tq--
PROFESSOR TODA: Precisely, and thank you so much
for your help.
So you have R10 of t, 1/2 sine of 2t
and t cubed over 3 minus e to the minus e plus 2.
And close, and that's it.
And box your answer.
So I got the long motion back.
Similarly, you could find, if somebody gives you
the acceleration of a long motion and asks you
this is the acceleration.
And I give you some initial values.
And you have to find first the velocity,
going backwards one step.
And from the velocity, backwards a second step,
get the position vector.
And that sounds a little bit more elaborate.
But it doesn't have to be a long computation.
In general, we do not focus on giving you
an awfully long computation.
We just want to test your understanding of the concepts.
And having this in mind, I picked another example.
I would like to see what that is.
And the initial velocity will be given in this case.
This is what I was thinking a little bit ahead of that.
So somebody gives you the acceleration in the velocity
vector at 0 and is asking you to find the velocity vector So
let me give it to you for t between 0 and 2 pi.
I give you the acceleration vector,
it will be nice and sassy.
Let's see, that's going to be cosine of t, sine of t and 0.
And you'll say, oh, I know how to do those.
Of course you know.
But I want you to pay attention to the constraints
of integration.
This is why I do this kind of exercise again.
So what do we have for V of t.
V of 0 is-- somebody will say, let's give something nice,
and let's say this would be-- I have no idea what I want.
Let's say i, j, and that's it.
How do you do that?
V of t.
Let's integrate together.
You don't like this?
I hope that by now, you've got used to it.
A bracket, I'm doing a bracket, like in the book.
So sine t plus a constant.
What's the integral of sine, class?
V equals sine t plus a constant.
And C3 is a constant.
And there I go.
You say, oh my god, what am I having?
V of 0-- is as a vector, I presented it
in the canonical standard basis as 1, 1, and 0.
So from that one, you can jump to this one
and say, yes, I'm going to plug in 0, see what I get.
In the general formula, when you plug in 0,
you get C1-- what is cosine of 0?
Minus 1, I have here, plus C2.
And C3, that is always there.
And then V of 0 is what I got here.
V of 0 has to be compared to what your initial data was.
So C1 is 1, C2 is 2, and C3 is--
So let me replace it.
I say the answer will be-- cosine t plus 1, sine t plus 2,
and the constants.
But then somebody, who is really an experimental guy,
says well--
STUDENT: You have it backwards.
It's sine of t plus 1, and then you
have the cosine of t plus 2.
PROFESSOR TODA: Oh, yeah.
Wait a minute.
This is-- I miscopied looking up.
So I have sine t, I was supposed to-- minus cosine t
and I'm done.
So thank you for telling me.
So sum t plus 1 minus cosine t plus 2 and 0
are the functions that I put here by replacing C1, C2, C3.
And then, somebody says, wait a minute,
now let me give you V of 0.
Let me give you R of 0.
We have zeroes already there.
And you were supposed to get R from here.
So what is R of t, the position vector, find it.
V of t is given.
Actually, it's given by you, because you found it
at the previous step.
And R of 0 is given as well.
And let's say that would be-- let's say 1, 1, and 1.
So what do you need to do next?
You have R prime given.
That leaves you to integrate to get R t.
And R of t is going to be what?
Who is going to tell me what I have to write down?
Minus cosine t plus t plus-- let's use the constant K1
integration.
And then what?
STUDENT: Sine of t.
PROFESSOR TODA: I think it's minus sine, right?
Minus sine of t plus 2t plus K2 and K3, right?
So R of 0 is going to be what?
First of all, we use this piece of information.
Second of all, we identify from the formula we got.
So from the formula I got, just plugging in 0,
it should come out straight as minus 1 plus K1.
0 for this guy, 0 for the second term, K2 and K3.
So who is helping me solve the system really quickly?
K1 is 2.
K2 is--
STUDENT: 1.
PROFESSOR TODA: K3 is 1.
And I'm going back to R and replace it.
And that's my final answer for this two-step problem.
So I have a two-step integration from the acceleration
to the velocity, from the velocity
to the position vector.
Minus cosine t plus t plus 2.
Remind me, because I have a tendency to miscopy,
an I looking in the right place?
Yes.
So I have minus sine t plus 2t plus 1 and K3 is one.
So this is the process you are supposed to remember
for the rest of the semester.
It's not a hard one.
It's something that everybody should master.
Is it hard?
How many of you understood this?
Please raise hands.
Oh, no problem, good.
Now would you tell me-- I'm not going to ask you
what kind of motion this is.
It's a little bit close to a circular motion but not
a circular motion.
However, can you tell me anything interesting
about the type of trajectory that I have, in terms
of the acceleration vector?
The acceleration vector is beautiful,
just like in the case of the washer.
That was a vector that-- like this
would be the circular motion.
The acceleration would be this unique vector
that comes inside.
Is this going outside or coming inside?
Is it a unit vector?
Yes, it is a unit vector.
So suppose that I'm looking at the trajectory,
if it were more or less a motion that has
to do with mixing into a bowl.
Would this go inside or outside?
Towards the outside or towards the inside?
I plugged j-- depends on what I'm looking at, in terms
of surface that I'm on, right?
Do you remember from last time we
had that helix that was on a cylinder.
And we asked ourselves, how is that [INAUDIBLE] pointing?
And it was pointing outside of the cylinder,
in the direction towards the outside.
Coming back to the review, there are
several things I'd like to review but not all of them.
Because some of the examples we have there,
you understood them really well.
I was very proud of you, and I saw
that you finished-- almost all of you
finished the homework number one.
So I was looking outside at homework number
two that is over these three sections.
So I was hoping you would ask me today, between two and three,
if you have any difficulties with homework two.
That's due February 11.
And then the latest homework that I posted yesterday, I
don't know how many of you logged in.
But last night I posted a homework
that is getting a huge extended deadline, which
is the 28th of February.
Because somebody's birthday is February 29.
I was just thinking why would somebody need be a whole month?
You would need the whole month to have a good view
of the whole chapter 11.
I sent you the videos for chapter 11.
And for chapter 11, you have this huge homework
which is 49 problems.
So please do not, do not leave it
to the last five days or six days,
because it's going to kill you.
There are people who say, I can finish
this in the next five days.
I know you can.
I know you can, I don't doubt it.
That's why I left you so much freedom.
But you have-- today is the second or the third?
So practically you have 25 days to work on this.
On the 28th at 11 PM it's going to close.
I would work a few problems every other day.
Because I need a break, so I would alternate.
But don't leave it-- even if you have help,
especially if you have help, like a tutor or tutoring
services here that are free in the department.
Do not leave it to the last days.
Because you're putting pressure on yourself, on your brain,
on your tutor, on everybody.
Yes sir.
STUDENT: So that's homework three?
PROFESSOR TODA: That's homework three,
and it's a huge homework over chapter 11.
STUDENT: You said there are 49 problems?
PROFESSOR TODA: I don't remember exactly but 47, 49.
I don't remember how many.
STUDENT: Between 45 and 50.
PROFESSOR TODA: Between 45 and 50, yes.
If you encounter any bug-- although there shouldn't
be bugs, maybe 1 in 1,000.
If you encounter any bug that the programmer
of those problems may have accidentally put in,
you let me know.
So I can contact them.
If there is a problem that I consider shouldn't be there,
I will eliminate that later on.
But hopefully, everything will be doable,
everything will be fair and you will be able to solve it.
Any questions?
Particular questions from the homework?
STUDENT: [INAUDIBLE] is it to parametrize a circle of a set,
like of a certain radius on the xy-plane?
PROFESSOR TODA: Shall we do that?
Do you want me to do that in general, in xy-plane, OK.
STUDENT: [INAUDIBLE] in the xy-plane.
PROFESSOR TODA: xy-plane and then what was the equation?
Was it like a equals sine of t or a equals sine of bt?
Because it's a little bit different,
depending on how the parametrization was given.
What's your name again, I forgot.
I don't know what to refer you.
STUDENT: Ryder.
PROFESSOR TODA: Was that part of what's due on the 11th?
STUDENT: It doesn't-- yes, it doesn't give a revision set.
It says--
PROFESSOR TODA: Let me quickly read-- find parametrization
of the circle of radius 7 in the xy-plane, centered at 3, 1,
oriented counterclockwise.
The point 10, 1 should be connected--
STUDENT: Just one more second.
PROFESSOR TODA: Do you mind if I put it.
I'll take good care of it.
I won't drop it.
So the point-- parametrization of the circle of radius
7 in the xy-plane, centered at 3, 1.
So circle centered at-- and I'll say it x0, 1 0, being 3, 1.
No, because then I'm solving your problem.
But I'm solving your problem anyway,
even if I change change the numbers.
Why don't I change the numbers, and then
you do it for the given numbers.
Let's say 1, 0.
And it's the same type of problem, right?
Oriented counterclockwise.
That's important.
So you have circle radius 7.
I think people could have any other,
because problems are-- sometimes you get a random assignment.
So you have R equals 2, let's say.
And you have the point, how to make up something.
The point corresponding to t equals
0 will be given as you have [INAUDIBLE], 1, 0, whatever.
OK?
Use the t as the parameter for all your answers.
So use t as a parameter for all your answers,
and the answers are written in the interactive field as x of t
equals what and y of t equals what,
and it's waiting for you to fill them in.
You know.
OK, now I was talking to [INAUDIBLE].
I'm going to give this back to you.
Thank you, Ryan.
So when you said it's a little bit frustrating,
and I agree wit you, that in this variant of webwork
problems you have to enter both of them correctly
in order to say yes, correct.
I was used to another library-- the library was outdated
[INAUDIBLE]-- where if I enter this correctly I get 50%
credit, and if I enter this incorrectly it's not going
to penalize me.
So I a little bit complained about it,
and I was shown the old library where
I can go ahead and go back and assign problems
where you get the answer correct for this one
and incorrect for this one, and you get partial credit.
So I'm probably going to switch to that.
Let's do that.
This is a very good problem.
I'm glad you brought it up.
What have you learned about conics in high school?
You've learned about-- well, it depends.
You've learned about ellipse.
You've learned about hyperbola.
You've learned about parabola.
Some of you put them down for me for extra credit.
I was very happy you did that.
It's a good exercise.
If you have-- Alex, yes?
STUDENT: I was just thinking, does that say 1, 0?
The point corresponding to t0 [INAUDIBLE]?
PROFESSOR TODA: I think that's what I meant.
I don't know, I just came up with it.
I made it.
1, 0.
I make up all my problems.
STUDENT: But the center of the circle isn't 1, 0.
PROFESSOR TODA: Oh, oops.
Yes.
Sorry.
So 2, 0.
No--
[INTERPOSING VOICES]
PROFESSOR TODA: --because the radius.
This is the problem when you don't think very [INAUDIBLE].
I always like to make up my own problems.
When an author, when we came up with the problems in the book,
of course we had to think, draw, and make sure they made sense.
But when you just come up with a problem out of the middle
of nowhere-- thank you so much.
Of course, we would have realized
that was nonsense in just a minute.
But it's good that you told me.
So x of t, y of t.
Let's find it.
Based on what?
What is the general equation of a circle?
x minus x0 squared plus y minus y0 squared equals R squared.
And you have learned that in high school.
Am I right or not?
You have.
OK.
Good.
Now, in our case what is x0 and what is y0?
x0 is 1 and y0 is 0.
Because that's why-- I don't know.
I just made it up.
And I said that's the center.
I'll draw.
I should have drawn it in the beginning,
and that would have helped me not come up
with some nonsensical data.
c is 1, 0.
Radius is 2.
So I'm going this way.
What point is this way, guys?
Just by the way.
Because [INAUDIBLE] is 1, 0, right?
And this way the other extreme, the antipode is 3, 0.
So that's exactly what Alexander was saying.
And now it makes sense.
Well, I cannot draw today.
STUDENT: [INAUDIBLE]
PROFESSOR TODA: It looks horrible.
It looks like an egg that is shaped-- disabled egg.
OK.
All right.
So the motion of-- the motion will come like that.
From t equals 0, when I'm here, counterclockwise,
I have to draw-- any kind of circle you have in the homework
should be drawn on the board.
If you have a general, you don't know what the data is.
I want to help you solve the general problem.
For the original problem, which is a circle
of center x, 0, y, 0 and radius R, generic one,
what is the parametrization without data?
Without specific data.
What is the parametrization?
And I want you to pay attention very well.
You are paying attention.
You are very careful today.
[INAUDIBLE]
So what do you have?
STUDENT: Cosine.
PROFESSOR TODA: Before that cosine
there is an R, excellent.
So [INAUDIBLE] there R cosine of t.
I'm not done.
What do I put here?
STUDENT: Over d.
PROFESSOR TODA: No, no.
I'm continuing.
STUDENT: Plus x0.
PROFESSOR TODA: Plus x0.
And R sine t plus y0.
Who taught me that?
First of all, this is not unique.
It's not unique.
I could put sine t here and cosine t here
and it would be the same type of parametrization.
But we usually put the cosine first
because we look at the x-axis corresponding
to the cosine and the y-axis corresponding to the sine.
If I don't know that, because I happen to know that
from when I was 16 in high school, if I don't know that,
what do I know?
I cook up my own parametrization.
And that's a very good thing.
And I'm glad Ryan asked about that.
How does one come up with this?
Do we have to memorize?
In mathematics, thank god, we don't memorize much.
The way we cook up things is just from, in this case,
from the Pythagorean theorem of-- no.
Pythagorean theorem of trigonometry?
The fundamental identity of trigonometry,
which is the same thing as the Pythagorean theorem.
What's the fundamental identity of trigonometry?
Cosine squared plus sin squared equals 1.
If I have a problem like that, I must
have that this is R cosine t and this is R sine t.
Because when I take the red guys and I
square them and I add them together,
I'm going to have R squared.
All righty, good.
So no matter what kind of data you have,
you should be able to come up with this on your own.
And what else is going to be happening?
When I solve for x of-- the point corresponding to t
equals 0.
x of 0 and y of 0 will therefore be what?
It will be R plus x0.
This is going to be what?
Just y0.
Does anybody give them to me?
STUDENT: 3, 0.
PROFESSOR TODA: Alexander gave me the correct ones.
They will be 3 and 0.
Are you guys with me?
They could be anything, anything that makes sense.
All right, for example somebody would say, I'm starting here.
I give you other points.
Then you put them in, you plug in that initial point,
meaning that you're starting your motion here.
And you do go around the circle one
because, you take [INAUDIBLE] only between 0 and 2 pi.
Alexander.
STUDENT: I have [INAUDIBLE].
PROFESSOR TODA: OK.
STUDENT: [INAUDIBLE]
PROFESSOR TODA: No, I thought that I misprinted something
again.
STUDENT: No, I was about to say something really dumb.
PROFESSOR TODA: OK.
So how do we make sense of what we have here?
Well, y0 corresponds to what I said.
So this is a superfluous equation.
I don't need that.
What do I know from that?
R will be 2.
x1 is 1.
I have a superfluous equation.
I have to get identities in that case, right?
OK, now.
What is going to be my-- my bunch of equations
will be x of t equals 2 cosine t plus 1 and y of t
equals-- I don't like this marker.
I hate it.
Where did I get it?
In the math department.
And it's a new one.
I got it as a new one.
It's not working.
OK, y of t.
The blue contrast is invisible.
I have 2 sine t.
Okey dokey.
When you finish a problem, always quickly
verify if what you got makes sense.
And obviously if I look at everything,
it's matching the whole point.
Right?
OK.
Now going back to-- this is reminding me of something in 3d
that I wanted to talk to you today about.
This is [INAUDIBLE].
I'm going to give you, in a similar way
with this simple problem, I'm going
to give you something more complicated
and say find the parametrization of a helix.
And you say, well, I'm happy that this
isn't a made-up problem again.
I have to be a little bit more careful
with these made-up problems so that they make sense.
Of a helix R of t such that it is contained or it lies,
it lies on the circular cylinder x squared
plus y squared equals 4.
Why is that a cylinder?
The z's missing, so it's going to be a cylinder whose
main axis is the z axis.
Right?
Are you guys with me?
I think we are on the same page.
And you cannot solve the problem just with this data.
Do you agree with me?
And knowing that, the curvature of the helix is k
equals 2/5 at every point.
And of course it's an oxymoron.
Because what I proved last time is
that the curvature of a helix is a constant.
So remember, we got the curvature of a helix
as being a constant.
STUDENT: What's that last word of the sentence?
It's "the curvature is at every" what?
PROFESSOR TODA: At every point.
I'm sorry I said, it very-- I abbreviated [INAUDIBLE].
So at every point you have the same curvature.
When you draw a helix you say, wait,
the helix is bent uniformly.
If you were to play with a spring taken from am old bed,
you would go with your hands along the spring.
And then you say, oh, it bends about the same.
Yes, it does.
And that means the curvature is the same.
How would you solve this problem?
This problem is hard, because you cannot integrate
the curvature.
Well, what is the curvature?
The curvature would be--
STUDENT: Absolute value.
PROFESSOR TODA: Just absolute value of R
double prime if it were in s.
And you cannot integrate.
If somebody gave you the vector equation
of double prime of this, them you say,
yes, I can integrate one step going back.
I get R prime of s.
Then I go back to R of s.
But this is a little bit complicated.
I'm giving you a scalar.
You have to be a little bit aware of what you did last time
and try to remember what we did last time.
What did we do last time?
I would not give you a problem like that
on the final, because it would assume that you have solved
the problem we did last time in terms of R of t
equals A equals sine t.
A sine t and [? vt. ?]
And we said, this is the standard parametrized helix
that sits on a cylinder of radius A and has the phb.
So the distance between consecutive spirals
really matters.
That really makes the difference.
STUDENT: I have a question.
PROFESSOR TODA: You wanted to ask me something.
STUDENT: Is s always the reciprocal of t?
Are they always--
PROFESSOR TODA: No, not reciprocal.
You mean s of t is a function is from t0 to t of the speed.
R prime and t-- d tau, right?
Tau not t. [INAUDIBLE].
t and s are different parameters.
Different times.
Different parameter times.
And you say--
STUDENT: Isn't s the parameter time
when [INAUDIBLE] parametrized?
PROFESSOR TODA: Very good.
So what is the magic s?
I'm proud of you.
This is the important thing to remember.
t could be any time.
I start measuring wherever I want.
I can set my watch to start now.
It could be crazy.
Doesn't have to be uniform.
Motion, I don't care.
s is a friend of yours that says,
I am that special time so that according to me
the speed will become one.
So for a physicist to measure the speed with respect to this,
parameter s time, the speed will always become one.
That is the arclength time and position.
How you get from one another, I told you last time
that for both of them you have-- this is R of t
and this is little r of s.
And there is a composition.
s can be viewed as a function of t,
and t can be viewed as a function of s.
As functions they are inverse to one another.
So going back to who they are, a very good question,
because this is a review anyway, [? who wants ?]
s as a function of t for this particular problem?
I hope you remember, we were like-- have you seen this movie
with Mickey Mouse going on a mountain that
was more like a cylinder.
And this is the train going at a constant slope.
And one of my colleagues, actually, he's at Stanford,
was telling me that he gave his students in Calc 1
to prove, formally prove, that the angle formed
by the law of motion by the velocity vector,
with the horizontal plane passing through the particle,
is always a constant.
I didn't think about doing in now, but of course we can.
We could do that.
So maybe the next thing would be, like,
if you [INAUDIBLE] an extra problem, can
we show that angle between the velocity vector on the helix
and the horizontal plane through that point is a constant.
STUDENT: Wouldn't it just be, because B of t
is just a constant times t?
PROFESSOR TODA: Yeah.
We'll get to that.
We'll get to that in a second.
So he reminded me of an old movie from like 70 years ago,
with Mickey Mouse and the train.
And the train going up at the same speed.
You have to maintain the same speed.
Because if you risk it not, then you sort of
are getting trouble.
So you never stop.
If you stop you go back.
So it's a regular curve.
What I have here is that such a curve.
Regular curve, never stop.
Get up with a constant speed.
Do you guys remember the speed from last time?
We'll square root the a squared plus b squared.
When we did the velocity thingie.
And I get square root a squared plus b squared times t.
Now, today I would like to ask you one question.
What if-- Ryan brought this up.
It's very good.
b is a constant.
What if b would not be a constant,
or maybe could be worse?
For example, instead of having another linear function with t,
but something that contains higher powers of t.
Then you don't go at the constant speed anymore.
You can say goodbye to the cartoon.
Yes, sir?
STUDENT: And then it's [INAUDIBLE].
One that goes [INAUDIBLE].
PROFESSOR TODA: I mean, it's still--
STUDENT: s is not multiplied by a constant.
The function between t and s is not a constant one.
PROFESSOR TODA: It's going to be a different parameterization,
different speed.
Sometimes-- OK, you have to understand.
Let's say I have a cone.
And I'm going slow at first, and I
go faster and faster and faster and faster
to the end of the cone.
But then I have the same physical curve,
and I parameterized [INAUDIBLE] the length.
And I say, no, I'm a mechanic.
Or I'm the engineer of the strain.
I can make the motion have a constant speed.
So even if the helix is no longer circular,
and it's sort of a crazy helix going on top of the mountain,
as an engineer I can just say, oh no,
I want cruise control for my little train.
And I will go at the same speed.
See, the problem is the slope a constant.
And thinking of what they did that
stand for, because it didn't stand
for [INAUDIBLE] in honors.
We can do it in honors as well.
We'll do it in a second.
Now, k obviously is what?
Some of you have very good memory,
and like the memory of a medical doctor, which is great.
Some of you don't.
But if you don't you just go back and look at the notes.
What I'm trying to do, but I don't know,
it's also a matter of money-- I don't
want to use the math department copier-- I'd
like to make a stack of notes.
So that's why I'm collecting these notes, to bring them back
to you.
For free!
I'm not going to sell them to you.
I'm [INAUDIBLE].
So that you can have those with you whenever you want,
or put them in a spiral, punch holes in them,
and have them for review at any time.
Reminds me of what that was-- that was in the notes.
a over a squared plus b squared.
So who can tell me, a and b really quickly,
so we don't waste too much time, Mr. a is--?
STUDENT: So this is another way
STUDENT: 2.
PROFESSOR TODA: 2.
STUDENT: So is this another way of defining k in k of s?
PROFESSOR TODA: Actually--
STUDENT: That's the general curvature for [INAUDIBLE].
PROFESSOR TODA: You know what is the magic thing?
Even if-- the curvature is an invariant.
It doesn't depend the reparametrization.
There is a way maybe I'm going to teach you, although this
is not in the book.
What are the formulas corresponding
to the [INAUDIBLE] t and v that depend on curvature and torsion
and the speed along the curve.
And if you analyze the notion of curvature, [INAUDIBLE],
no matter what your parameter will be, t, s, tau,
God knows what, k will still be the same number.
So k is viewed as an invariant with respect
to the parametrization.
STUDENT: So then that a over a squared plus b squared,
that's another way of finding k?
PROFESSOR TODA: Say it again?
STUDENT: So using a over a squared plus b squared
is another way of finding k?
PROFESSOR TODA: No.
Somebody gave you k.
And then you say, if it's a standard parametrization,
and then I get 2/5, can I be sure a is 2?
I'm sure a is 2 from nothing.
This is what makes me aware that a is 2 the first place.
Because its the radius of the cylinder.
This is x squared, x and y.
You see, x squared plus y squared is a squared.
This is where I get a from.
a is 2.
I replace it in here and I say, all righty,
so I only have one choice. a is 2 and b is?
STUDENT: [INAUDIBLE]
PROFESSOR TODA: But can b plus-- So what I'm saying,
a is 2, right?
We know that from this.
If I block in here I have 4 and somebody says plus minus 1.
No.
b is always positive.
So you remember the last time we discussed
about the standard parametrization.
But somebody will say, but what if I put a minus?
What if I'm going to put a minus?
That's an excellent question.
What's going to happen if you put minus t?
[INTERPOSING VOICES]
PROFESSOR TODA: Exactly.
In the opposite direction.
Instead of going up, you go down.
All right.
Now, I'm gonna-- what else?
Ah, I said, let's do this.
Let's prove that the angle is a constant,
the angle that's made by the velocity
vector of the train with the horizontal plane is a constant.
Is this hard?
Nah.
Yes, sir?
STUDENT: Are we still going to find R of t given only k?
PROFESSOR TODA: But didn't we?
We did.
R of t was 2 cosine t, 2 sine t, and t.
All right?
OK, so we are done.
What did I say?
I said that let's prove-- it's a proof.
Let's prove that the angle made by the velocity to the train--
to the train?-- to the direction of motion, which is the helix.
And the horizontal plane is a constant.
Is this hard?
How are we going to do that?
Now I start waking up, because I was very tired.
STUDENT: [INAUDIBLE]
PROFESSOR TODA: Excuse me.
STUDENT: [INAUDIBLE]
PROFESSOR TODA: So you see, the helix contains this point.
And I'm looking at the velocity vector
that is standard to the helix.
And I'll call that R prime.
And then you say, yea, but how am I
going to compute that angle?
What is that angle?
STUDENT: It's a function of b.
PROFESSOR TODA: It will be.
But we have to do it rigorously.
So what's going to happen for me to draw that angle?
First of all, I should take-- from the tip
of the vector I should draw perpendicular
to the horizontal plane passing through the point.
And I'll get P prime.
God knows why.
I don't know why, I don't know why. [? Q. ?] And this is PR,
and P-- not PR.
PR is too much [INAUDIBLE] radius, M.
OK, so then you would take PQ and then
you would measure this angle.
Well, you have to be a little bit smarter than that,
because you can prove something else.
This is the complement of another angle that you love.
And using chapter 9 you can do that angle in no time.
So this is the complement of the angle
formed by the velocity vector of prime with the normal.
But not the normal principle normal to the curve,
but the normal to the plane.
And what is the normal to the plane?
Let's call the principal normal n to the curve big N bar.
So in order to avoid confusion, I'll write this little n.
How about that?
Do you guys know-- like they do in mechanics.
If you have two normals, they call that 1n.
1 is little n, and stuff like that.
So this is the complement.
If I were able to prove that that complement
is a constant-- this is the Stanford [? property-- ?] then
I will be happy.
Is it hard?
No, for god's sake.
Who is little n?
Little n would be-- is that the normal to a plane
that you love?
What is your plane?
STUDENT: xy plane.
PROFESSOR TODA: Your plane is horizontal plane.
STUDENT: xy.
PROFESSOR TODA: Yes, xy plane.
Or xy plane shifted, shifted, shifted, shifted.
That's the normal change?
No.
Who is the normal?
STUDENT: [INAUDIBLE]
PROFESSOR TODA: [INAUDIBLE].
STUDENT: 0, 0, 1.
PROFESSOR TODA: 0, 0, 1.
OK.
When I put 0 I was [INAUDIBLE].
So this is k.
All right.
And what is our prime?
I was-- that was a piece of cake.
We did it last time minus a sine t, a equals sine t and b.
Let's find that angle.
Well, I don't know.
You have to teach me, because you have chapter 9 fresher
in your memory than I have it.
Are you taking attendance also?
Are you writing your name down?
Oh, no problem whatsoever.
STUDENT: We didn't get it.
PROFESSOR TODA: You didn't get it.
Circulate it.
All right, so who is going to help me with the angle?
What is the angle between two vectors, guys?
That should be review from what we just covered in chapter 9.
Let me call them u and v. And who's
going to tell me how I get that angle?
STUDENT: [INAUDIBLE] is equal to the inverse cosine of the dot
product of [? the magnitude. ?]
PROFESSOR TODA: Do you like me to write arc
cosine or cosine [INAUDIBLE].
Doesn't matter.
Arc cosine of--
STUDENT: The dot products.
PROFESSOR TODA: The dot product between u and v.
STUDENT: Over magnitude.
PROFESSOR TODA: Divided by the product of their magnitudes.
Look, I will change the order, because you're not
going to like it.
Doesn't matter.
OK?
So the angle phi between my favorite vectors
here is going to be simply the dot product.
That's a blessing.
It's a constant.
STUDENT: So you're doing the dot product
between the normal [INAUDIBLE]?
PROFESSOR TODA: Between this and that.
So this is u and this is v. So the dot product
would be 0 plus v. So the dot product
is arc cosine of v, which, thank god, is a constant.
I don't have to do anything anymore.
I'm done with the proof bit, because arc cosine
of a constant will be a constant.
OK?
All right.
So I have v over what?
What is the length of this vector?
1. [INAUDIBLE].
What's the length of that vector?
Square root of a squared plus b squared.
All right?
STUDENT: How did you [INAUDIBLE].
PROFESSOR TODA: So now let me ask you one thing.
What kind of function is arc cosine?
Of course I said arc cosine of a constant is a constant.
What kind of a function is arc cosine?
I'm doing review with you because I think it's useful.
Arc cosine is defined on what with values in what?
STUDENT: Repeat the question?
PROFESSOR TODA: Arc cosine.
Or cosine inverse, like Ryan prefers.
Cosine inverse is the same thing.
It's a function defined by where to where?
Cosine is defined from where to where?
From R to minus 1.
It's a cosine of t.
t could be any real number.
The range is minus 1, 1.
Close the interval.
STUDENT: So it's-- so I just wonder why--
PROFESSOR TODA: Minus 1 to 1, close interval.
But pay attention, please.
Because it cannot go back to R. It has to be a 1 to 1 function.
You cannot have an inverse function if you don't take
a restriction of a function to be 1 to 1.
And we took that restriction of a function.
And do you remember what it was?
[INTERPOSING VOICES]
PROFESSOR TODA: 0 to pi.
Now, on this one I'm really happy.
Because I asked several people-- people
come to my office to get all sorts of transcripts,
[INAUDIBLE].
And in trigonometry I asked one student,
so you took trigonometry.
So do you remember that?
He didn't remember that.
So I'm glad you do.
How about when I had the sine inverse?
How was my restriction so that would be a 1 to 1 function?
It's got to go from minus 1 to 1.
What is the range?
[INTERPOSING VOICES]
PROFESSOR TODA: Minus pi over 2.
You guys know your trig.
Good.
That's a very good thing.
You were in high school when you learned that?
Here at Lubbock High?
STUDENT: Yes.
PROFESSOR TODA: Great.
Good job, Lubbock High.
But many students, I caught them, who wanted credit
for trig who didn't know that.
Good.
So since arc cosine is a function that is of 0, pi,
for example, what if my-- let me give you an example.
What was last time, guys?
a was 1. b was 1.
For one example.
In that case, 1 with 5b.
STUDENT: [INAUDIBLE] ask you for the example you just did?
PROFESSOR TODA: No last time.
STUDENT: A was 3 and b was--
PROFESSOR TODA: So what would that be, in this case 5?
STUDENT: That would be b over the square root--
STUDENT: 3 over pi.
PROFESSOR TODA: a is 1 and b is 1, like we did last time.
STUDENT: [INAUDIBLE] 2, which is--
PROFESSOR TODA: Plug in 1 is a, b is 1.
What is this?
STUDENT: It's just pi over 4.
PROFESSOR TODA: Pi over 4.
So pi will be our cosine, of 1 over square root 2, which
is 45 degree angle, which is-- you said pi over 4, right?
[INAUDIBLE].
So exactly, you would have that over here.
This is where the cosine [INAUDIBLE].
Now you see, guys, the way we have, the way I assume a and b,
the way anybody-- the book also introduces
a and b to be positive numbers.
Can you tell me what kind of angle phi will be,
not only restricted to 0 pi?
Well, a is positive.
b is positive.
a doesn't matter.
The whole thing will be positive.
Arc cosine of a positive number--
STUDENT: Between 0 and pi over 2.
PROFESSOR TODA: That is.
Yeah, so it has to be between 0 and pi over 2.
So it's going to be only this quadrant.
Does that make sense?
Yes, think with the imagination of your eyes,
or the eyes of your imagination.
OK.
You have a cylinder.
And you are moving along that cylinder.
And this is how you turn.
You turn with that little train.
Du-du-du-du-du, you go up.
When you turn the velocity vector and you
look at the-- mm.
STUDENT: The normal.
PROFESSOR TODA: The normal!
Thank you.
The z axis, you always have an angle between 0 and pi over 2.
So it makes sense.
I'm going to go ahead and erase the whole thing.
So we reviewed, more or less, s of t, integration, derivation,
moving from position vector to velocity to acceleration
and back, acceleration to velocity to position vector,
the meaning of arclength.
There are some things I would like to tell you,
because Ryan asked me a few more questions about the curvature.
The curvature formula depends very
much on the type of formula you used for the curve.
So you say, wait, wait, wait, Magdelena,
you told us-- you are confusing us.
You told us that the curvature is uniquely
defined as the magnitude of the acceleration vector
when the law of motion is an arclength.
And that is correct.
So suppose my original law of motion was R of t [INAUDIBLE]
time, any time, t, any time parameter.
I'm making a face.
But then from that we switch to something beautiful,
which is called the arclength parametrization.
Why am I so happy?
Because in this parametrization the magnitude of the speed
is 1.
And I define k to be the magnitude
of R double prime of s, right?
The acceleration only in the arclength [? time ?]
parameterization.
And then this was the definition.
A. Can you prove-- what?
Can you prove the following formula?
T prime of s equals k times N of s.
This is famous for people who do-- not for everybody.
But imagine you have an engineer who does
research of the laws of motion.
Maybe he works for the railways and he's
looking at skew curves, or he is one
of those people who project the ski slopes,
or all sorts of winter sports slope or something, that
involve a lot of curvatures and torsions.
That guy has to know the Frenet formula.
So this is the famous first Frenet formula.
Frenet was a mathematician who gave the name to the TNB
vectors, the trihedron.
You have the T was what?
The T was the tangent [INAUDIBLE] vector.
The N was the principal unit normal.
In those videos that I'm watching that I also sent you--
I like most of them.
I like the Khan Academy more than everything.
Also I like the one that was made by Dr. [? Gock ?]
But Dr. [? Gock ?] made a little bit of a mistake.
A conceptual mistake.
We all make mistakes by misprinting or misreading
or goofy mistake.
But he said this is the normal vector.
This is not-- it's the principle normal vectors.
There are many normals.
There is only one tangent direction,
but in terms of normals there are many that
are-- all of these are normals.
All the perpendicular in the plane-- [INAUDIBLE]
so this is my law of motion, T. All this plane is normal.
So any of these vectors is a normal.
The one we choose and defined as T prime
over T prime [INAUDIBLE] absolute values
called the principal normal.
It's like the principal of a high school.
He is important.
So T and B-- B goes down, or goes-- down.
Well, yeah, because B is T cross N. So when you find the Frenet
Trihedron, TNB, it's like that.
T, N, and B. What's special, why do we call it the frame,
is that every [? payer ?] of vectors
are mutually orthogonal.
And they are all unit vectors.
This is the famous Frenet frame.
Now, Mr. Frenet was a smart guy.
He found-- I don't know whether he was adopting mathematics
or not.
Doesn't matter.
He found a bunch of formulas, of which this is the first one.
And it's called a first Frenet formula.
That's one thing I want to ask you.
And then I'm going to give you more formulas for curvatures,
depending on how you define your curve.
So for example, base B based on the definition one
can prove that for a curve that is not parametrizing
arclength-- you say, ugh, forget about parametrization
in arclength.
This time you're assuming, I want to know!
I'm coming to this because Ryan asked.
I want to know, what is the formula directly?
Is there a direct formula that comes
from here for the curvature?
Yeah, but it's a lot more complicated.
When I was a freshman, maybe a freshman or a sophomore,
I don't remember, when I was asked to memorize
that, I did not memorize it.
Then when I started working as a faculty member,
I saw that I am supposed to ask it from my students.
So this is going to be R prime plus product
R double prime in magnitude over R prime cubed.
So how am I supposed to remember that?
It's not so easy.
Are you cold there?
It's cold there.
I don't know how these roofs are made.
Velocity times acceleration.
This is what I try to teach myself.
I was old already, 26 or 27.
Velocity times acceleration, cross product,
take the magnitude, divide by the speed, cube.
Oh my god.
So I was supposed to know that when I was 18 or 19.
Now, I was teaching majors of mechanical engineering.
They knew that by heart.
I didn't, so I had to learn it.
So if one is too lazy or it's simply
inconvenient to try to reparametrize from R of T
being arclength parametrization R of s and do that thing here,
one can just plug in and find the curvature like that.
For example, guys, as Ryan asked,
if I have A cosine, [INAUDIBLE], and I do this with respect
to T, can I get k without-- k will not
depend on T or s or tau.
It will always be the same.
I will still get A over A squared plus B
squared, no matter what.
So even if I use this formula for my helix,
I'm going to get the same thing.
I'll get A over A squared plus B squared,
because curvature is an invariant.
There is another invariant that's-- the other invariant,
of course, in space is called torsion.
We want to talk a little bit about that later.
So is this hard?
No.
It shouldn't be hard.
And you guys should be able to help me on that, hopefully.
How do we prove that?
STUDENT: N is G prime [INAUDIBLE].
PROFESSOR TODA: That's right, proof.
And that's a very good start, wouldn't you say?
So what were the definitions?
Let me start from the definition of T.
That's going to be-- I am in hard planes, right?
So you say, wait, why do you write it as a quotient?
You're being silly.
You are in arclength, Magdalena.
I am.
I just pretend that I cannot see that.
So if I'm in arclength, that means
that the denominator is 1.
So I'm being silly.
So R prime of s is T. Say it again.
R prime of s is T. OK.
Now, did we know that T and N are orthogonal?
How did we know that T and N were orthogonal?
We proved that last time, actually.
T and N are orthogonal.
How do I write that? [INAUDIBLE].
Meaning that T is perpendicular to N, right?
From the definition.
You said it right, Sandra.
But why is it from the definition
that I can jump to conclusions and say, oh,
since I have T prime here, then this is perpendicular to T?
Well, we did that last time.
STUDENT: Two parallel vectors.
PROFESSOR TODA: We did it-- how did we do it?
We did this last.
We said T dot T equals 1.
Prime the whole thing.
T prime times T plus T times T prime, T dot T prime will be 0.
So T and T prime are perpendicular always.
Right?
OK, so the whole thing is a colinear vector to T prime.
It's just T prime times the scalar.
So he must be perpendicular to T.
So T and N are perpendicular.
So I do have the direction of motion.
I know that I must have some scalar here.
How do I prove that this scalar is the curvature?
So if I have-- if they are colinear-- why are
they colinear?
T perpendicular to T prime implies that T prime
is colinear to N. Say it again.
If T and T prime are perpendicular to one another,
that means T prime is calling it to the normal.
So here I may have alph-- no alpha.
I don't know!
Alpha over [INAUDIBLE] sounds like a curve.
Give me some function.
STUDENT: u of s?
PROFESSOR TODA: Gamma of s.
u of s, I don't know.
So how did I conclude that?
From T perpendicular to T prime.
Now from here on, you have to tell me why
gamma must be exactly kappa.
Well, let's take T prime from here.
T prime from here will give me what?
T prime is our prime prime.
Say what?
Our prime prime.
What is our prime prime?
Our [? problem ?] prime of s.
STUDENT: You have one too many primes inside.
PROFESSOR TODA: Oh my god.
Yeah.
So R prime prime.
So T prime in absolute value will
be exactly R double prime of s.
Oh, OK.
Note that from here also T prime of s in absolute value,
in magnitude, I'm sorry, has to be gamma of s.
Why is that?
Because the magnitude of N is 1.
N is unique vector by definition.
So these two guys have to coincide.
So R double prime, the best thing
that I need to do, it must coincide
with the scalar gamma of s.
So who is the mysterious gamma of s?
He has no chance but being this guy.
But this guy has a name.
This guy, he's the curvature [? cap ?] of s by definition.
Remember, Ryan, this is the definition.
So by definition the curvature was the magnitude
of the acceleration in arclength.
OK.
Both of these guys are T prime in magnitude.
So they must be equal from here and here.
It implies that my gamma must be kappa.
And I prove the formula.
OK.
How do you say something is proved?
Because this is what we wanted.
We wanted to replace this generic scalar function
to prove that this is just the curvature.
QED.
That's exactly what we wanted to prove.
Now, whatever scalar function you have here,
that must be the curvature.
Very smart guy, this Mr. Frenet.
I'm now going to take a break.
If you want to go use the bathroom really quickly,
feel free to do it.
I'm just going to clean the board,
and I'll keep going in a few minutes.
STUDENT: [INAUDIBLE]
PROFESSOR TODA: I will do it-- well,
actually I want to do a different example, simple one,
which is a plain curve, and show that the curvature has a very
pretty formula that you could [INAUDIBLE] memorize,
that in essence is the same.
But it depends on y equals f of x.
[INAUDIBLE] So if somebody gives you
a plane called y equals f of x, can you
write that curvature [INAUDIBLE] function of f?
And you can.
And again, I was deep in that when I was 18 or 19
as a freshman.
But unfortunately for me I didn't learn it at that time.
And several years later when I started teaching engineers,
well, they are mostly mechanical.
And mechanical engineering [INAUDIBLE].
They knew those, and they needed those in every research paper.
So I had to learn it together with them.
I'll worry about [INAUDIBLE].
STUDENT: Can you do a really ugly one, like [INAUDIBLE]?
PROFESSOR TODA: I can do some ugly ones.
And once you know the general parametrization,
it will give you a curvature.
Now I'm testing your memory.
Let's see what you remember.
Um-- don't look at the notes.
A positive function, absolute-- actually,
magnitude of what vector?
STUDENT: R prime.
PROFESSOR TODA: R prime velocity plus acceleration speed cubed.
Right?
OK.
Now, can we take advantage of what we just learned
and find-- you find with me, of course, not
as professor and student, but like a group of students
together.
Let's find a simple formula corresponding
to the curvature of a plane curve.
And the plane curve could be [INAUDIBLE]
in two different ways, just because I want
you to practice more on that.
Either given as a general parametrization-- guys,
what is the general parametrization
I'm talking about for a plane curve?
x of t, y of t, right?
x equals x of t.
y equals y of t.
So one should not have to do that all the time,
not have to do that for a simplification like a playing
card.
We have to find another formula that's pretty, right?
Well, maybe it's not as pretty.
But when is it really pretty?
I bet it's going to be really pretty if you have a plane
curve even as you're used to in an explicit form--
I keep going.
No stop. [INAUDIBLE].
I think it's better.
We make better use of time this way.
Or y equals f of x.
This is an explicit way to write the equation of a curve.
OK, so what do we need to do?
That should be really easy.
R of t being the first case of our general parametrization,
x equals x of t, y equals y of t will be-- who tells me, guys,
that-- this is in your hands.
Now you convinced me that, for whatever reason,
you [INAUDIBLE].
You became friends with these curves.
I don't know when.
I guess in the process of doing homework.
Am I right?
I think you did not quite like them before or the last week.
But I think you're friends with them now.
x of t, y of t.
Let people talk.
STUDENT: 0.
PROFESSOR TODA: So.
Great.
And then R prime of t will be x prime of t, y prime of t,
and 0.
I assume this to be always non-zero.
I have a regular curve.
R double prime will be-- x double prime where
double prime-- we did the review today
of the lasting acceleration.
Now, your friends over here, are they nice or mean?
I hope they are not so mean.
The cross product is a friendly fellow.
You have i, j, k, and then the second row
would be x prime, y prime, 0.
The last row would be x double prime, y double prime, 0.
And it's a piece of cake.
OK, piece of cake, piece of cake.
But I want to know what the answer is.
So you have exactly 15 seconds to answer this question.
Who is R prime plus R double prime as a [? coordinate. ?]
[INTERPOSING VOICES]
PROFESSOR TODA: Good.
x prime, y double prime minus x double prime, y prime times k.
And it doesn't matter when I take the magnitude,
because magnitude of k is 1.
So I discovered some.
This is how mathematicians like to discover new formulas based
on the formulas they [? knew. ?] They
have a lot of satisfaction.
Look what I got.
Of course, they in general have more complicated things to do,
and they have to check and recheck.
But every piece of a computation is a challenge.
And that gives people satisfaction.
And when they make a mistake, it brings a lot of tears as well.
So what-- could be written on the bottom, what's
the speed cubed?
Speed is coming from this guy.
So the speed of the velocity, the magnitude of the velocity
is the speed.
And that-- going to give you square.
I'm not going to write down [INAUDIBLE].
Square root of x squared, x prime squared times
y prime squared, and I cube that.
Many people, and I saw that in engineering, they
don't like to put that square root anymore.
And they just write x prime squared plus y prime squared
to the what power?
STUDENT: 3/2.
PROFESSOR TODA: 3/2.
So this is very useful for engineering styles,
when you have to deal with plane curves, motions
in plane curves.
But now what do you have in the case,
in the happy case, when you have y equals f of x?
I'm going to do that in a second.
I want to keep this formula on the board.
What's the simplest parametrization?
Because that's why we need it, to look over
parametrizations again and again.
R of t for this plane curve will be-- what is t?
x is t, right?
x is t, y is f of t.
Piece of cake.
So you have t and f of t.
And how many of you watched the videos that I sent you?
Do you prefer Khan Academy, or do you
prefer the guys, [INAUDIBLE] guys who are lecturing?
The professors who are lecturing in front of a board or in front
of a-- what is that?
A projector screen?
I like all of them.
I think they're very good.
I think you can learn a lot from three
or four different instructors at the same time.
That's ideal.
I guess that you have this chance only now
in the past few years.
Because 20 years ago, if you're didn't like your instructor
or just you couldn't stand them, you had no other chance.
There was no YouTube, no internet,
no way to learn from others.
R prime of t would be 1 f prime of t.
But instead of t I'll out x, because x is t.
I don't care.
R double prime of t would be 0, f double prime of x.
So I feel that, hey, I know what's going to come up.
And I'm ready.
Well, we are ready to write it down.
This is going to be Mr. x prime.
This is going to be replacing Mr. y prime.
This is going to replace Mr. a double prime.
This is going to be replacing Mr. y double prime of x.
Oh, OK, all right.
So k, our old friend from here will become what?
And I'd better shut up, because I'm talking too much.
STUDENT: [INAUDIBLE] double prime [INAUDIBLE].
PROFESSOR TODA: That is the absolute value, mm-hmm.
[? n ?] double prime of x, and nothing else.
Right, guys?
Are you with me?
Divided by--
STUDENT: [INAUDIBLE]
PROFESSOR TODA: Should I add square root?
I love square roots.
I'm crazy about them.
So you go 1 plus f prime squared cubed.
So that's going to be-- any questions?
Are you guys with me?
That's going to be the formula that I'm going
to use in the next example.
In case somebody wants to know-- I got
this question from one of you.
Suppose we get a parametrization of a circle
in the midterm or the final.
Somebody says, I have x of t, just like we did it
today, a cosine t plus 0.
And y of t equals a sine t plus y 0.
What is this, guys?
This is a circle, a center at 0, y, 0, and radius a.
Can use a better formula-- that anticipated my action today--
to actually prove that k is going to be [? 1/a? ?]
Precisely.
Can we do that in the exam?
Yes.
So while I told you a long time ago
that engineers and mathematicians observed
hundreds of years ago-- actually,
somebody said, no, you're not right.
The Egyptians already saw that.
They had the notion of inverse proportionality
in Egypt, which makes sense if you look at the pyramids.
So one look at the radius, it says if the radius is 2,
then the curvature is not very bent.
So the curvature's inverse proportion [INAUDIBLE]
the radius.
So if this is 2, we said the curvature's 1/2.
If you take a big circle, the bigger
the radius, the smaller the bending
of the arc of the circle, the smaller of the curvature.
Apparently the ancient world knew that already.
They Egyptians knew that.
The Greeks knew that.
But I think they never formalized it--
not that I know.
So if you are asked to do this in any exam,
do you think that would be a problem?
Of course we would do review.
Because people are going to forget this formula, or even
the definition.
You can compute k for this formula.
And we are going to get k to the 1/a.
This is a piece of cake, actually.
You may not believe me, but once you plug in the equations
it's very easy.
Or you can do it from the definition
that gives you k of s.
You'll reparametrize this in arclength.
You can do that as well.
And you still get 1/a.
The question that I got by email,
and I get a lot of email.
I told you, that keeps me busy a lot,
about 200 emails every day.
I really like the emails I get from students,
because I get emails from all sorts of sources--
Got some spam also.
Anyway, what I'm trying to say, I got this question last time
saying, if on the midterm we get such a question,
can we say simply, curvature's 1/a, a is the radius.
Is that enough?
Depends on how the problem was formulated.
Most likely I'm going to make it through that or show that.
Even if you state something, like, yes, it's 1/a,
with a little argument, it's inverse proportional
to the radius, I will still give partial credit.
For any argument that is valid, especially
if it's based on empirical observation,
I do give some extra credit, even if you didn't
use the specific formula.
Let's see one example.
Let's take y equals e to the x.
No, let's take e to the negative x.
Doesn't matter.
y equals e to the negative x.
And let's make x between 0 and 1.
I'll say, write the curvature.
Write the equation or the formula of the curvature.
And I know it's 2 o'clock and I am answering questions.
This was a question that one of you had during the short break
we took.
Can we do such a problem?
Like she said.
Yes, I [INAUDIBLE] to the negative
x because I want to catch somebody
not knowing the derivative.
I don't know why I'm doing this.
Right?
So if I were to draw that, OK, try and draw that, but not now.
Now, what formula are you going to use?
Of course, you could do this in many ways.
All those formulas are equivalent for the curvature.
What's the simplest way to do it?
Do y prime.
Minus it to the minus x.
Note here in this problem that even if you mess up and forget
the minus sign, you still get the final answer correct.
But I may subtract a few points if I see something nonsensical.
y double prime equals--
[INTERPOSING VOICES]
--plus e to the minus x.
And what is the curvature k of t?
STUDENT: y prime over--
PROFESSOR TODA: Oh, I didn't say one more thing.
I want the curvature, but I also want the curvature
in three separate moments, in the beginning, in the end,
and in the middle.
STUDENT: Don't we need to parametrize it
so we can [INAUDIBLE] x prime [INAUDIBLE]?
PROFESSOR TODA: No.
Did I erase it?
STUDENT: Yeah, you did.
PROFESSOR TODA: [INAUDIBLE].
And one of my colleagues said, Magda, you are smart,
but you are like one of those people who,
in the anecdotes about math professors,
gets out of their office and starts walking
and stops a student.
Was I going this way or that way?
And that's me.
And I'm sorry about that.
I should not have erased that.
I'm going to go ahead and rewrite it,
because I'm a goofball.
So the one that I wanted to use k of x will be f double prime.
STUDENT: And cubed.
PROFESSOR TODA: Cubed!
Thank you.
So that 3/2, remember it, [INAUDIBLE] 3/2 [INAUDIBLE]
square root cubed.
Now, for this one, is it hard?
No.
That's a piece of cake.
I said I like it in general, but I also
like it-- find the curvature of this curve in the beginning.
You travel on me.
From time 0 to 1 o'clock, whatever.
One second.
That's saying this is in seconds to make it more physical.
I want the k at 0, I want k at 1/2, and I want k at 1.
And I'd like you to compare those values.
And I'll give you one more task after that.
But let me start working.
So you say you help me on that.
[INAUDIBLE]
Minus x over square root of 1 plus--
STUDENT: [INAUDIBLE]
PROFESSOR TODA: Right.
So can I write this differently, a little bit differently?
Like what?
I don't want to square each of the minus 2x.
Can I do that?
And then the whole thing I can say to the 3/2
or I can use the square root, whichever is your favorite.
Now, what is k of 0?
STUDENT: 0.
Or 1.
PROFESSOR TODA: Really?
STUDENT: 1/2.
3/2.
PROFESSOR TODA: So let's take this slowly.
Because we can all make mistakes, goofy mistakes.
That doesn't mean we're not smart.
We're very smart, right?
But it's just a matter of book-keeping and paying
attention, being attentive.
OK.
When I take 0 and replace-- this is drying fast.
I'm trying to draw it.
I have 1 over 1 plus 1 to the 3/2.
I have a student in one exam who was just-- I don't know.
He was rushing.
He didn't realize that he had to take it slowly.
He was extremely smart, though.
1 over-- you have that 1 plus 1 is 2.
2 to the 1/2 would be square root of 2 cubed.
It would be exactly 2 square root of 2.
And more you can write this as rationalized.
Now, I have a question for you.
[INAUDIBLE]
I'm When we were kids, if you remember-- you are too young.
Maybe you don't remember.
But I remember when I was a kid, my teacher would always ask me,
rationalize your answer.
Rationalize your answer.
Put the rational number in the denominator.
Why do you think that was?
For hundreds of years people did that.
STUDENT: [INAUDIBLE]
PROFESSOR TODA: Because they didn't have a calculator.
So we used to, even I used to be able to get the square root out
by hand.
Has anybody taught you how to compute square root by hand?
You know that.
Who taught you?
STUDENT: I don't remember it.
My seventh grade teacher taught us.
PROFESSOR TODA: There is a technique
of taking groups of twos and then fitting the--
and they still teach that.
I was amazed, they still teach that
in half of the Asian countries.
And it's hard, but kids in fifth and sixth grade
have that practice, which some of us learned and forgot about.
So imagine that how people would have done this, and of course,
square root of 2 is easy.
1.4142, blah blah blah.
Divide by 2.
You can do it by hand.
At least a good approximation.
But imagine having a nasty square root there to compute,
and then you would divide by that natural number.
You have to rely on your own computation to do it.
There were no calculators.
How about k of 1?
How is that?
What is that?
e to the minus 1.
That's a little bit harder to compute, right?
1 plus [INAUDIBLE].
What is that going to be?
Minus 2.
Replace it by 1 to the 3/2.
I would like you to go home and do the following.
[INAUDIBLE]-- Not now, not now.
We stay a little bit longer together.
k of 0, k of 1/2, and k of 1.
Which one is bigger?
And one last question about that, how much extra credit
should I give you?
One point?
One point if you turn this in.
Um, yeah.
Four, [? maybe ?] two points.
Compare all these three values, and find
the maximum and the minimum of kappa of t,
kappa of x, for the interval where
x is in the interval 0, 1.
0, closed 1.
Close it.
Now, don't ask me, because it's extra credit.
One question was, by email, can I ask my tutor to help me?
As long as your tutor doesn't write down your solution,
you are in good shape.
Your tutor should help you understand some constants,
spend time with you.
But they should not write your assignment themselves.
OK?
So it's not a big deal.
Not I want to tell you one secret that I normally don't
tell my Calculus 3 students.
But the more I get to know you, the more
I realize that you are worth me telling you about that.
STUDENT: [INAUDIBLE]
PROFESSOR TODA: No.
There is a beautiful theory that engineers
use when they start the motions of curves and parametrizations
in space.
And that includes the Frenet formulas.
And you already know the first one.
And I was debating, I was just reviewing what I taught you,
and I was happy with what I taught you.
And I said, they know about position vector.
They know about velocity, acceleration.
They know how to get back and forth from one another.
They know our claim.
They know how to [? reparameterize our ?]
claims.
They know the [INAUDIBLE] and B. They know already
the first Frenet formula.
They know the curvature.
What else can I teach them?
I want to show you-- one of you asked me,
is this all that we should know?
This is all that a regular student should know in Calculus
3, but there is more.
And you are honor students.
And I want to show you some beautiful equations here.
So do you remember that if I introduce r of s
as a curving arclength, that is a regular curve.
I said there is a certain famous formula that is T prime of s
called-- leave space.
Leave a little bit of space.
You'll see why.
It's a surprise.
k times-- why don't I say k of s?
Because I want to point out that k is an invariant.
Even if you have another parameter,
would be the same function.
But yes, as a function of s, would be k times N bar, bar.
More bars because they are free vectors.
They are not bound to a certain point.
They're not married to a certain point.
They are free to shift by parallelism in space.
However, I'm going to review them as bound at the point
where they are.
So they-- no way they are married to the point
that they belong to.
Maybe the [? bend ?] will change.
I don't know how it's going to change like crazy.
Something like that.
At every point you have a T, an N, and it's a 90 degree angle.
Then you have the binormal, which makes a 90 degree
angle-- [INAUDIBLE].
So the way you should imagine these corners
would be something like that, right?
90-90-90.
It's just hard to draw them.
Between the vectors you have-- If you draw T and N, am I
right, that is coming out?
No.
I have to switch them.
T and N. Now, am I right?
Now I'm thinking of the [? faucet. ?]
If I move T-- yeah, now it's coming out.
So this is not getting into the formula.
So this is the first formula.
You say, so what?
You've taught that.
We proved it together.
What do you want from us?
I want to teach you two more formulas.
N prime.
And I'd like you to leave more space here.
So you have like an empty field here and an empty field here
[INAUDIBLE].
If you were to compute T prime, the magic thing
is that T prime is a vector.
N prime is a vector.
B prime is a vector.
They're all vectors.
They are the derivatives of the vectors T and NB.
And you say, why would I care about the derivatives
of the vectors T and NB?
I'll tell you in a second.
So if you were to compute in prime,
you're going to get here.
Minus k of s times T of s.
Leave room.
Leave room, because there is no component that
depends on N. No such component that that depends on N.
This is [INAUDIBLE].
There is nothing in N. And then in the end
you'll say, plus tau of s times B. There is missing--
something is.
And finally, if you take B prime,
there is nothing here, nothing here.
In the middle you have minus tau of s times N of s.
And now you know that nobody else but you knows that.
The other regular sections don't know these formulas.
What do you observe about this bunch of equations?
Say, oh, wait a minute.
First of all, why did you put it like that?
Looks like a cross.
It is a cross.
It is like one is shaped in the name of the Father, of the Son,
and so on.
So does it have anything to do with religion?
No.
But it's going to help you memorize better the equations.
These are the famous Frenet equations.
You only saw the first one.
What do they represent?
If somebody asks you, what is k?
What it is k of s?
What's the curvature?
You go to a party.
There are only nerds.
It's you.
Some people taking advanced calculus or some people
from Physics, and they say, OK, have you heard of the Frenet
motion, Frenet formulas, and you say,
I know everything about it.
What if they ask you, what is the curvature of k?
You say, curvature measures how a curve is bent.
And they say, yeah, but the Frenet formula tells you
more about that.
Not only k shows you how bent the curve is.
But k is a measure of how fast T changes.
And he sees why.
Practically, if you take the [INAUDIBLE] to the bat,
this is the speed of T. So how fast the teaching will change.
That will be magnitude, will be just k.
Because magnitude of N is 1.
So note that k of s is the length of T prime.
This measures the change in T. So how fast T varies.
What does the torsion represent?
Well, how fast the binormal varies.
But if you want to think of a helix,
and it's a little bit hard to imagine,
the curvature measures how bent a certain curve is.
And it measures how bent a plane curve is.
For example, for the circle you have radius a, 1/a, and so on.
But there must be also a function that
shows you how a curve twists.
Because you have not just a plane curve where
you care about curvature only.
But in the space curve you care how the curves twist.
How fast do they move away from a certain plane?
Now, if I were to draw-- is it hard to memorize these?
No.
I memorized them easily based on the fact
that everything looks like a decomposition
of a vector in terms of T, N, and B. So in my mind
it was like, I take any vector I want, B. And this is T,
this is N, and this is B. Just the weight was IJK.
Instead if I, I have T. Instead of J, I have N. Instead of K,
I have B. They are still unit vectors.
So locally at the point I have this frame
and I have any vector.
This vector-- I'm a physicist.
So let's say I'm going to represent that as v1 times
the T plus v2 times-- instead of J,
we'll use that N plus B3 times-- that's
the last element of the bases.
Instead of k I have v. So it's the same here.
You try to pick a vector and decompose
that in terms of T, N, and B. Will I put that on the final?
No.
But I would like you to remember it, especially
if you are an engineering major or physics major,
that there is this kind of Frenet frame.
For those of you who are taking a-- for differential equations,
you already do some matrices and built-in systems
of equations, systems of differential equations.
I'm not going to get there.
But suppose you don't know differential equations,
but you know a little bit of linear algebra.
And I know you know how to multiply matrices.
You know how I know you multiply matrices,
no matter how much mathematics you learn.
And most of you, you are not in general algebra this semester.
Only two of you are in general algebra.
When I took a C++ course, the first homework I got was
to program a matrix multiplication.
I have to give in matrices.
I have to program that in C++.
And freshmen knew that.
So that means you know how to write this as a matrix
multiplication.
Can anybody help me?
So T, N, B is the magic triple.
T, N, B's the magic corner.
T, N, and B are the Three Musketeers who are all
orthogonal to one another.
And then I do derivative with respect to s.
If I want to be elegant, I'll put d/ds.
OK.
How am I going to fill in this matrix?
So somebody who wants to know about differential equations,
this would be a--
STUDENT: 0, k, 0.
PROFESSOR TODA: Very good.
0, k, 0, minus k 0 tau, 0 minus tau 0.
This is called the skew symmetric matrix.
Such matrices are very important in robotics.
If you've ever been to a robotics team,
like one of those projects, you should
know that when we study motions of-- let's say
that my arm performs two rotations in a row.
All these motions are described based
on some groups of rotations.
And if I go into details, it's going to be really hard.
But practically in such a setting
we have to deal with matrices that either have determined
one, like all rotations actually have,
or have some other properties, like this guy.
What's the determinant of this guy?
What do you guys think?
Just look at it.
STUDENT: 0?
PROFESSOR TODA: 0.
It has determinant 0.
And moreover, it looks in the mirror.
So this comes from a group of motion,
which is little s over 3, the linear algebra, actually.
So when k is looking in the mirror,
it becomes minus k tau, is becoming minus tau.
It is antisymmetric or skew symmetric.
Skew symmetric or antisymmetric is the same.
STUDENT: Antisymmetric, skew symmetric matrix.
PROFESSOR TODA: Skew symmetric or antisymmetric
is exactly the same thing.
They are synonyms.
So it looks in the mirror and picks up the minus sign,
has 0 in the bag.
What am I going to put here?
You already got the idea.
So when Ryan gave me this, he meant
that he knew what I'm going to put here, as a vector,
as a column vector.
STUDENT: [INAUDIBLE]
PROFESSOR TODA: No, no no.
How do I multiply?
TNB, right?
So guys, how do you multiply matrices?
You go first row and first column.
So you go like this.
0 times T plus k times 10 plus 0 times B. Here it is.
So I'm teaching you a little bit more
than-- if you are going to stick with linear algebra
and stick with differential equations,
this is a good introduction to more of those mathematics.
Yes, sir?
STUDENT: Why don't you use Cramer's rule?
PROFESSOR TODA: Uh?
STUDENT: Why don't you use the Cramer's rule?
PROFESSOR TODA: The Cramer's rule?
STUDENT: Yeah. [INAUDIBLE].
PROFESSOR TODA: No.
First of all, Crarmer's rule is to solve systems of equations
that don't involve derivatives, like a linear system
like Ax equals B. I'm going to have,
for example, 3x1 plus 2x3 equals 1.
5x1 plus x2 plus x3 equals something else.
So for that I can use Cramer's rule.
But look at that!
This is really complicated.
It's a dynamical system.
At every moment of time the vectors are changing.
So it's a crazy [INAUDIBLE].
Like A of t times something, so some vector
that is also depending on time equals the derivative
of that vector that [INAUDIBLE].
So that's a OD system that you should learn in 3351.
So I don't know what your degree plan is,
but most of you in engineering will
take my class, 2316 in algebra, OD1 3350 where they teach you
about differential equations.
These are all differential equations, all three of them.
In 3351 you learn about this system
which is a system of differential equation.
And then you practically say, now I
know everything I need to know in math, and you say,
goodbye math.
If you guys wanted to learn more,
of course I would be very happy to learn that, hey, I
like math, I'd like to be a double major.
I'd like to be not just an engineering, but also math
major if you really like it.
Many people already have a minor.
Many of you have a minor in your plan.
Like for that minor you only need--
STUDENT: One extra math course.
PROFESSOR TODA: One extra math course.
For example, with 3350 you don't need 3351 for a minor.
Why?
Because you are taking the probability in stats anyway.
You have to.
They force you to do that, 3342.
So if you take 3351 it's on top of the minor that we give you.
I know because that's what I do.
I look at the degree plans.
And I work closely to the math adviser, with Patty.
She has all the [INAUDIBLE].
STUDENT: So is [INAUDIBLE]?
PROFESSOR TODA: You mean double?
Double degree?
We have this already in place.
We've had it for many years.
It's an excellent plan.
162 hours it is now.
It used to be 159.
Double major, computer science and mathematics.
And I could say they were some of the most successful
in terms of finding jobs.
What would you take on top of that?
Well, as a math major you have a few more courses
to take one top of that.
You can link your computer science with the mathematics,
for example, by taking numerical analysis.
If you love computers and you like calculus
and you want to put together all the information
you have in both, then numerical analysis
would be your best bet.
And they require that in both computer science degree
if you are a double major, and your math degree.
So the good thing is that some things count for both degrees.
And so with those 160 hours you are very happy.
Oh, I'm done, I got a few more hours.
Many math majors already have around 130.
They're not supposed to.
They are supposed to stop at 120.
So why not go the extra 20 hours and get two degrees in one?
STUDENT: It's a semester.
PROFESSOR TODA: Yeah.
Of course, it's a lot more work.
But we have people who like-- really they
are nerdy people who loved computer science from when
they were three or four.
And they also like math.
And they say, OK, I want to do both.
OK, a little bit more and I'll let you go.
Now I want you to ask me other questions
you may have had about the homework, anything that
gave you headache, anything that you feel you need a little bit
more of an explanation about.
Yes?
STUDENT: I just have one.
In WeBWork, what is the easiest way
to take the square root of something?
STUDENT: sqrt.
PROFESSOR TODA: sqrt is what you type.
But of course you can also go to the caret 1/2.
Something non-technical?
Any question, yes sir, from the homework?
Or in relation to [INAUDIBLE]?
STUDENT: I don't understand why is the tangent unit vector,
it's just the slope off of that line, right?
The drunk bug?
Whatever line the drunk bug is on?
PROFESSOR TODA: So it would be the tangent
to the directional motion, which is a curve.
And normalized to have length one.
Because otherwise our prime is-- you may say,
why do you need T to be unitary?
OK, computations become horrible unless your speed
is 1 or 5 or 9.
If the speed is a constant, everything else becomes easier.
So that's one reason.
STUDENT: And why is the derivative
of T then perpendicular?
Why does it always turn into--
PROFESSOR TODA: Perpendicular to T?
We've done that last time, but I'm glad to do it again.
And I forgot what we wrote in the book,
and I also saw in the book this thing
that if you have R, in absolute value, constant--
and I've done that with you guys--
prove that R and R prime had every point perpendicular.
So if you have-- we've done that before.
Now, what do you do then?
T [INAUDIBLE] T is 1.
The scalar [INAUDIBLE] the product.
T prime times T plus T prime T prime.
So 0.
And T is perpendicular to T prime,
because that means T or T prime equals 0.
When you run in a circle, you say--
OK, let's run in a circle.
I say, this is my T. I can feel that there is something that's
trying to bend me this way.
That is my acceleration.
And I have to-- but I don't know-- how familiar are you
with the winter sports?
In many winter sports, the Frenet Trihedron is crucial.
Imagine that you have one of those slopes,
and all of the sudden the torsion becomes too weak.
That means it becomes dangerous.
That means that the vehicle you're in,
the snow vehicle or any kind of-- your skis, [INAUDIBLE],
if the torsion of your body moving can become too big,
that will be a problem.
So you have to redesign that some more.
And this is what they do.
You know there have been many accidents.
And many times they say, even in Formula One,
the people who project a certain racetrack,
like a track in Indianapolis or Montecarlo
or whatever, they have to have in mind
that Frenet frame every second.
So there are simulators showing how
the Frenet frame is changing.
There are programs that measure the curvature
in a torsion for those simulators at every point.
Neither the curvature nor the torsion
can exceed a certain value.
Otherwise it becomes dangerous.
You say, oh, I thought only the speed is a danger.
Nope.
It's also the way that the motion, if it's a skew curve,
it's really complicated.
Because you twist and turn and bend in many ways.
And it can become really dangerous.
Speed is not [INAUDIBLE].
STUDENT: So the torsion was the twists in the track?
PROFESSOR TODA: The torsion is the twist.
And by the way, keep your idea.
You wanted to ask something more?
When you twist-- suppose you have something like a race car.
And the race car is at the walls of the track.
And here's-- when you have a very abrupt curvature
and torsion, and you can have that in Formula One as well,
why do they build one wall a lot higher than the other?
Because the poor car-- I don't know how passionate you
are about Formula One or car races--
the poor car is going to be close to the wall.
It's going to bend like that, that wall would be round.
And as a builder, you have to build the wall really high.
Because that kind of high speed, high velocity,
high curvature, the poor car's going szhhhhh-- then
again on a normal track.
Imagine what happens if the wall is not high enough.
The wheels of the car will go up and get over.
And it's going to be a disaster.
So that engineer ha to study all the parametric equations
and the Frenet frame and deep down make a simulator,
compute how tall the walls should be in order for the car
not to get over on the other side or get off the track.
It's really complicated stuff.
It's all mathematics and physics,
but all the applications are run by engineers and-- yes, sir?
STUDENT: What's the difference [INAUDIBLE] centrifugal force?
PROFESSOR TODA: The centrifugal force
is related to our double prime.
Our double prime is related to N and T at the same time.
So at some point, let me ask you one last question and I'm done.
What's the relationship between acceleration or double prime?
And are they the same thing?
And when are they not the same thing?
Because you say, OK, practically the centrifugal--
STUDENT: They're the same on a curve.
PROFESSOR TODA: They are the same--
STUDENT: Like on a circle.
PROFESSOR TODA: On a circle!
And you are getting so close.
It's hot, hot, hot.
On a circle and on a helix they are the same up to a constant.
So what do you think the magic answer will be?
N was what, guys?
N was-- remind me again.
That was T prime over absolute value of T prime.
But that doesn't mean, does not equal, in general,
does not equal to R double prime.
When is it equal?
In general it's not equal.
When is it equal?
If you are in aclength, you see the advantage of aclength.
It's wonderful.
In arclength, T is R prime of s.
And in arclength that means T prime is R double prime of s.
And in arclength I just told you,
T prime is the first Frenet formula.
It'll be curvature times the N.
So the acceleration practically and the N
will be the same in arclength, up to a scalar multiplication.
But what if your speed is not even constant?
Then God help you.
Because the acceleration R double prime and N
are not colinear.
So if I were to draw-- and that's my last picture--
let me give you a wild motion here.
You start slow and then you go crazy and fast and slow down.
Just like most of the physical models from the bugs
and the flies and so on.
In that kind of crazy motion you have a T and N at every point.
[INAUDIBLE]
[? v ?] will be down.
And T is here.
So can you draw arc double prime for me?
It will still be towards the inside.
But it's still going to coincide with N. Maybe this one.
What's the magic thing is that T, N, and R double prime
are in the same plane always.
That's another secret other students
don't know in Calculus 3.
That same thing is called osculating plane.
We have a few magic names for these things.
So T and N, the plane that is-- how shall I say that?
I don't know.
The plane given by T and N is called osculating plane.
The acceleration is always on that plane.
So imagine T and N are in the same shaded plane.
R double prime is in the same plane.
OK?
Now, can you guess the other two names?
So this is T, this is N. And B is up.
This is my body's direction.
T and N, look at me.
T, N, and B. I'm the Frenet Trihedron.
Which one is the osculating plane?
It's the horizontal xy plane.
OK, do you know-- maybe you're a mechanical engineering major,
and after that I will let you go.
No extra credit, though for this task.
Maybe I'm going to start asking questions and give you $1.
I used to do that a lot in differential equations,
like ask a hard question, whoever gets it first,
give her a dollar.
Until a point when they asked me to teach Honors 3350 when
I started having three or four people answering the question
at the same time.
And that was a significant expense,
because I had to give $4 away at the same time.
STUDENT: I feel like you should've just
split it between--
PROFESSOR TODA: So that's normal and binormal.
This is me, the binormal, and this is the normal.
Does anybody know the name of this plane,
between normal and bionormal?
This would be this plane.
STUDENT: The skew [INAUDIBLE].
PROFESSOR TODA: Normal and binormal.
They call that normal plane.
So it's tricky if you are not a mechanical engineering major.
But some of you are maybe and will learn that later.
Any other questions for me?
Now, in my office I'm going to do review.
I was wondering if you have time,
I don't know if you have time to come to my office,
but should you have any kind of homework related question,
I'll be very happy to answer it now.
3:00 to 5:00.
Now, one time I had a student who
only had seven questions left.
He came to my office and he left with no homework.
We finished all of them.
And I felt guilty.
But at the same, he said, well, no, it's
better I came to you instead of going to my tutor.
It was fine.
So we can try some problems together today
if you want between 3:00 and 5:00, if you have the time.
Some of you don't have the time.
All right?
If you don't have the time today,
and you would like to be helped [INAUDIBLE],
click Email Instructor.
I'm going to get the questions [INAUDIBLE].
You're welcome to ask me anything
at any time over there.
[CLASSROOM CHATTER]
PROFESSOR TODA: I have somebody who's taking notes.
STUDENT: Yeah, I know.
And that's why I was like--
PROFESSOR TODA: He's going to make a copy
and I'll give you a copy.
STUDENT: Yeah.
My Cal 1 teacher, Dr. [INAUDIBLE].
STUDENT: Thank you.
PROFESSOR TODA: Yes, yeah.
Have a nice day.
STUDENT: --got really mad when I don't take notes.
Because he felt like I was not, I guess--